When $n$ standard six-sided dice are rolled, the probability of obtaining a sum of $1994$ is $>0$ and is the same as the probability of obtaining a sum of $S$. What is the smallest possible value of $S$?
This question has been asked before on MSE, but I had one doubt that wasn't in any of the answers.
To obtain the least sum $S$ we need the least number of dices. For the least number of dices $n$ to give the sum of throws as $1994$, we need all the throws to be $6$ or as close to it. As $1994$ is not divisible by $6$ we can either take $332$ rolls giving $6$ and $1$ roll giving $2$$(332×6+1×2=1992+2=1994)$ or $329$ rolls giving $6$ and $4$ rolls giving $5$$(329×6+5×4=1974+20=1994)$
Now this is where I have a problem. In the second approach, to get the least sum we would need $4$ rolls to give $2$ and rest to give $1$ for least sum $S$
and to satisfy having same probability of occuring. But in the first approach we only need $1$ of the rolls to be $2$ and rest all $1$ which obviously gives a smaller sum.
However all the answers went with the latter approach. What am I missing?
Best Answer
I'm not sure I understand your approach. I'll just explain otherwise.
Since $1994$ is nothing special, let's take the case of $n=3$ dice. Sum ranges from $3$ to $18$.
Note that $P(3)=P(18)$ as only one way for either {$(1,1,1)$ and $(6,6,6)$}.
Then $P(4)=P(17)$. Since $3$ ways to obtain either. {$(1,1,2),(1,2,1),(2,1,1)$} and {$(6,6,5),(6,5,6),(5,6,6)$}
Observe that $S=4$ can be obtained from $S=3$ by incrementing one of the $1$'s. And $S=17$ can be obtained from $S=18$ by decrementing one of the $6$'s. Also, the number of ways of incrementing $=$ number of ways of decrementing.
This is precisely the symmetry the second answer in your link points out.
Do you see now why the answer should be $337$?